Congruences anв matrices

[Lucky]

Please, help me!
1)
The property 4 from book Cracking GRE (Chapter 6) says that if

a=b (mod n) is equivalent to a=b, b+(c-1)n (mod cn)

but I don't understand why?

2) A is a matrix with integer entries, det(A)=1, => the entries of the inverse of A will also be integers

why?

[Lucky]

I've already understood the 2nd statement.

If A consists of all integers and is nonsingular (detA=1), then inverse of A= adj(A)*inv(detA)).

adjA has integer entries, because A does, so, provided detA=1, inverse of A will contain only integers.

harray!

[lime]

1) Apparently, it is mistake in the book, since it is quite easy to find counterexample to the statement given in the book. I can just assume that S.Leduc implied next thing.

If a = b (mod n)

then for every integer "c"

a = b + 2n (mod 2n)
a = b + 3n (mod 3n)
...
a = b + (c-1)n (mod (c-1)n)
a = b + cn (mod cn).

[Lucky]

Tnx, I thought it is a misprint but was not sure.

[zombie]

I thought that must be an error also.

Other errors I found:

The chain rule example is completely wrong. If I let F(u,v,y) = z, u = f(v,y) and v = g(x,u), then I get the book's answer. Otherwise, there must have been a typo. Does that cohere with what others got?

Also, in Chapter 1, Cos(a+b) = cos a * cos b - (not +) sin a *sin b.

Were there any other glaring errors?

[lime]

I couldn't find example you're talking about. Can you provide number of example, page and your book's edition number?

Also, in Chapter 1, Cos(a+b) = cos a * cos b - (not +) sin a *sin b.

Yes this is a typo. But you've written it correctly here.

cos(a+b) = cosa*cosb - sina*sinb
cos(a-b) = cosa*cosb + sina*sinb

These are correct, while in the book signs in both eq. for cosine are opposite.

Btw. in some (I have really seen it) books (3 ed.) there is one more typo on the same page a little bit above.

sin(-a) = -sin(a)
cos(-a) = cos(a).